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Ch6.6 MappingCh6.7 Conformal Mapping

講者： 許永昌 老師

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ContentsConformal MappingMappings

TranslationRotationInversion

Branch Points and Multivalent Functions

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Conformal mapping ( 請預讀 P368~P370)

Mapping: zwThe mapping is conformal if angle and sense

of rotation are preserved by the mapping. If w=f(z) is analytic in a region R of the z-plane,

then the mapping of R onto its image in the w-plane is conformal, except at points where f ’(z)=0. Proof:

0analytic ' 0

arg arg arg arg ,

Therefore,

lim arg arg arg ' .

f i

f i f if i

zf z

f z f zww w z z

z z z

w z f z

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Conformal mapping (continue)Based on Cauchy-Riemann conditions, we get

2u=0=2v, u v=0.

They are orthogonal to each other The curves u=constant and v=constant are orthogonal to each other.

Example:w=z2=(x2-y2)+2ixyCode: z2_uv.m

y=sqrt(x2-u), y=v/(2x) Contour u= x2-y2 , v=2xy. -2 -1 0 1 2

-3

-2

-1

0

1

2

3

z-plane

-8 -6 -4 -2 0 2 4

-10

-8

-6

-4

-2

0

2

4

6

8

10

w-plane

Proper rotation

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Conformal Mapping (final)The mapping of w=z2.

From these figures, you will find that the contour lines of y=C and y=-C are the same in w-plane.

Reason: z=reiq and z’=rei q +ip . z2=r2ei2q=z’ 2.

Therefore, it has a two-to-one correspondence.

-2 -1 0 1 2

-3

-2

-1

0

1

2

3

z-plane

-4 -2 0 2 4-8

-6

-4

-2

0

2

4

6

8w-plane

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Mappings ( 請預讀 P360~P363)

Linear Transformation:Translation:

w=z+z0.

Rotation: w=cz=(rrc)ei(q+q

c).

Nonlinear Transformation:Inversion:

w=1/z=1/r e-iq.…

Code: mappings.m想像 w=z‘ 與 z 畫在同一個座標系

-4 -2 0 2 4 6

0

2

4

6

translation

-6 -4 -2 0 2 4

-2

0

2

4

rotation

-4 -2 0 2 4-3

-2

-1

0

1

2

inversion

-5 0 5 10 15 20

-10

-5

0

5

10

square

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ExerciseProve that w=1/z will map a straight line in z-

plane into a circle cross w=0.Try to add “z=z*(1+1i);x=real(z);y=imag(z);”

into the code mappings.m to see the result.

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Multivalent functions and Branch Points ( 請預讀 P363~P367)

Multivalent function:w=f(z), however, w is not unique for each z.

Example:w=sqrt(z)

If z=rei q ’=rei q +i2mp, w=?w=ln(z)

If z=rei q ’=reiq +i2mp, w=?

STOP TO THINK:Since

how can we say that f ’(z) does exist? analytic?

0

' lim and is a multivalent function,z

f z z f zf z f z

z

Hint: Restrict the allowed range of q’.

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Multivalent functions and Branch Points (continue)The cut line here joins the two branch point

singularities at 0 and .Dm: 2pm < q ’ < 2p (m+1 ).Sm : 2pm = q ’

Based on Morera’s theorem, we can get f(z) is analytic in zS0D0S1D1… SnDn.sqrt(z): n=1 and S0 = S2.ln(z): n=.We call this surface a Riemann surface.

* 莊 ( 土斤 ) 泰，張南岳，復變函數

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Multivalent functions and Branch Points (final)We can think that:

It is the basis for the entire calculus of residues.

Stop to think: Hint: Riemann surface.

200

00

1ln 2 .

i i

i

r e

r eCdz z iz

? Does it conflict with Morera's theorem?Czdz

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Homework6.6.26.6.36.6.56.6.66.6.76.7.16.7.4

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NounsConformal mapping:

P368Riemann surface: P366